Derivative Of A Square Root | EMathZone

Finding a derivative of the square roots of a function can be done by using derivative by definition or the first principle method.

Consider a function of the form $$y = \sqrt x $$.

First we take the increment or small change in the function. \[\begin{gathered} y + \Delta y = \sqrt {x + \Delta x} \\ \Rightarrow \Delta y = \sqrt {x + \Delta x} – y \\ \end{gathered} \]

Putting the value of function $$y = \sqrt x $$ in the above equation, we get \[ \Rightarrow \Delta y = \sqrt {x + \Delta x} – \sqrt x \]

Using the rationalizing method \[\begin{gathered}\Rightarrow \Delta y = \sqrt {x + \Delta x} – \sqrt x \times \frac{{\sqrt {x + \Delta x} + \sqrt x }}{{\sqrt {x + \Delta x} + \sqrt x }} \\ \Rightarrow \Delta y = \frac{{{{\left( {\sqrt {x + \Delta x} } \right)}^2} – {{\left( {\sqrt x } \right)}^2}}}{{\sqrt {x + \Delta x} + \sqrt x }} \\ \Rightarrow \Delta y = \frac{{x + \Delta x – x}}{{\sqrt {x + \Delta x} + \sqrt x }} \\ \Rightarrow \Delta y = \frac{{\Delta x}}{{\sqrt {x + \Delta x} + \sqrt x }} \\ \end{gathered} \]

Dividing both sides by $$\Delta x$$, we get \[\begin{gathered}\frac{{\Delta y}}{{\Delta x}} = \frac{{\Delta x}}{{\Delta x\left( {\sqrt {x + \Delta x} + \sqrt x } \right)}} \\ \frac{{\Delta y}}{{\Delta x}} = \frac{1}{{\left( {\sqrt {x + \Delta x} + \sqrt x } \right)}} \\ \end{gathered} \]

Taking the limit of both sides as $$\Delta x \to 0$$, we have \[\begin{gathered}\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{1}{{\sqrt {x + \Delta x} + \sqrt x }} \\ \frac{{dy}}{{dx}} = \frac{1}{{\sqrt {x + 0} + \sqrt x }} \\ \frac{{dy}}{{dx}} = \frac{1}{{2\sqrt x }}\\ \end{gathered} \]

NOTE: If we take any function in the square root function, then   \[\frac{{dy}}{{dx}} = \frac{1}{{2\sqrt {f\left( x \right)} }}\frac{d}{{dx}}f\left( x \right) = \frac{1}{{2\sqrt {f\left( x \right)} }}f’\left( x \right)\]

Example: Find the derivative of $$y = \sqrt {2{x^2} + 5}$$

We have the given function as \[y = \sqrt {2{x^2} + 5} \]

Differentiating with respect to variable $$x$$, we get \[\frac{{dy}}{{dx}} = \frac{d}{{dx}}\sqrt {2{x^2} + 5} \]

Now using the formula derivative of a square root, we have \[\begin{gathered}\frac{{dy}}{{dx}} = \frac{1}{{2\sqrt {2{x^2} + 5} }}\frac{d}{{dx}}\left( {2{x^2} + 5} \right) \\ \frac{{dy}}{{dx}} = \frac{{4x}}{{2\sqrt {2{x^2} + 5} }} \\ \frac{{dy}}{{dx}} = \frac{{2x}}{{\sqrt {2{x^2} + 5} }} \\ \end{gathered} \]